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## Karnataka 2nd PUC Maths Previous Year Question Paper June 2019

Time: 3 Hrs 15 Min

Max. Marks: 100

Part – A

Answer ALL the following questions: (10 × 1 = 10)

Question 1.

Let * be the binary operation on N given.by a * b = LCM of a and b, ∀ a ? b ∈ N.

Answer:

Given a * b = LCM of a and b ∀ a, b ∈ N 5*7 = 35

Question 2.

Find the principal value of cot^{-1}\(\left(-\frac{1}{\sqrt{3}}\right)\)

Answer:

Cot^{-1}\(\left(-\frac{1}{\sqrt{3}}\right)\) = π – Cot^{-1}\(\left(-\frac{1}{\sqrt{3}}\right)\)

[∵ cot^{-1}(-x) = π – cot^{-1}(x)]

= π – \(\frac{\pi}{3}\)

∴ cot^{-1} \(\left(\frac{-1}{\sqrt{3}}\right)=\frac{2 \pi}{3}\)

Question 3.

If A = [a_{ij}] where elements are given by a_{ij} = \(\frac { 1 }{ 2 }\)|-3i + j| construct 2 × 2 matrix.

Answer:

Given a_{ij} = \(\frac { 1 }{ 2 }\) |-3i + j|

a_{11} = \(\frac { 1 }{ 2 }\)|-3(1) + 1|= \(\frac { 1 }{ 2 }\)|-2| = 1

a_{12} = \(\frac { 1 }{ 2 }\) |-3(1) + 2| = \(\frac { 1 }{ 2 }\)|-l| = \(\frac { 1 }{ 2 }\)

a_{21} = \(\frac { 1 }{ 2 }\) |-3(2) + 2| = \(\frac { 1 }{ 2 }\) |-5| = \(\frac { 5 }{ 2 }\)

a_{22} = \(\frac { 1 }{ 2 }\) |-3(2) + 2| = \(\frac { 1 }{ 2 }\) |-4| = 2

Question 4.

Find a value of x for which

Answer:

3 – x^{2} = 3 – 8 ⇒ x^{2} = 8 ⇒ x = ±2\(\sqrt{2}\)

Question 5.

If y = cos^{-1} (e^{x}), find \(\frac{d y}{d x}\).

Answer:

y = cos^{-1} (e^{x}) (diff wrt x)

Question 6.

Find \(\int \sec ^{2}(7-4 x) d x\)

Answer:

Question 7.

If â = \(\frac{1}{\sqrt{14}}\) (2î + 3ĵ + k̂), then write the directions cosines of â.

Answer:

â = \(\frac{1}{\sqrt{14}}\) (2î + 3ĵ + k̂)

∴ Directions cosines are :

\(\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}\)

Question 8.

Find the intercepts cut off by the plane 2x + y – z = 5

Answer:

2x + y – z = 5 …..(1) ÷ 5

\(\frac{2 x}{5}+\frac{y}{5}-\frac{z}{5}=1\)

or \(\frac{x}{(5 / 2)}+\frac{y}{5}+\frac{z}{(-5)}=1\)

∴ x – intercept = \(\frac { 5 }{ 2 }\) ; y – intercept = 5, z – intercept = -5

Question 9.

Define feasible region in a linear programming problem.

Answer:

Feasible region is defined as the common region determined by all the constraints including non – negative constraints x ≥ 0, y ≥ 0 of a linear programming problem is called as Feasible region.

Question 10.

If P(A) = \(\frac { 3 }{ 5 }\) and P(B) = \(\frac { 1 }{ 5 }\) find P(A ∩ B), where A and B are independent events.

Answer:

P(A) = \(\frac { 3 }{ 5 }\); P(B) = \(\frac { 1 }{ 5 }\)

P(A ∩ B) = P(A).P(B)

[Since A and B are independent events]

= \(\frac{3}{5} \cdot \frac{1}{5}\) = \(\frac { 3 }{ 25 }\)

Part-B

Answer any TEN questions: (10 x 2 = 20)

Question 11.

Find g.f and f.g, if (x) = 8x^{3} and g(x) = x^{1/3}

Answer:

Given

f(x) = 8x^{3} ,g(x)= x^{1/3}

fog (x) = f(g(x))

= f(x^{1/3}) =8(x^{3})^{1/3} =8x

gof (x) = g(f(x)) = g(8x^{3})

= (8x^{3})^{1/3} = 2x

Question 12.

Prove that tan^{-1} \(\frac { 2 }{ 11 }\) + tan^{-1} \(\frac { 7 }{ 24 }\) = tan^{-1} \(\frac { 1 }{ 2 }\)

Answer:

Consider LHS : tan^{-1} (\(\frac { 2 }{ 11 }\)) + tan^{-1} (\(\frac { 7 }{ 24 }\))

Question 13.

Write cot^{-1} \(\left(\frac{1}{\sqrt{x^{2}-1}}\right)\) ,x > 1 in the simplest form.

Answer:

Let x = sec q ⇒ q = sec^{-1} x.

Now cot^{-1} \(\left(\frac{1}{\sqrt{x^{2}-1}}\right)\) = cot^{-1} \(\left(\frac{1}{\sqrt{\sec ^{2} \theta-1}}\right)\)

= cot^{-1} \(\left(\frac{1}{\tan \theta}\right)\) = cot^{-1}(cot θ) = θ = sec^{-1} x.

Question 14.

Find the area of the triangle with vertices (3,8), (-4, 2) and (5, 1) using determinants.

Answer:

Area of triangle

= \(\frac { 1 }{ 2 }\) [3(2 – 1) -8(-4 -5) + 1(-4 -10)]

= \(\frac { 1 }{ 2 }\) [3 + 72 -14] = \(\frac { 61 }{ 2 }\)sq.units

Question 15.

Find \(\frac{d y}{d x}\) if x^{2} + xy + y^{2} = 100.

Answer:

x^{2} + xy + y^{2} = 100 (diff wrt x)

2x + x.\(\frac{d y}{d x}\) + y.1 + 2y\(\frac{d y}{d x}\) = 0

x.\(\frac{d y}{d x}\) + 2y\(\frac{d y}{d x}\) = 2x – y

\(\frac{d y}{d x}\)(x + 2y) = -(2x + y)

\(\therefore \frac{d y}{d x}=\frac{-(2 x+y)}{x+2 y}\)

Question 16.

Find \(\frac{d y}{d x}\) if y = (logx)^{cosx}.

Answer:

Let y = (log x)^{cos x}

⇒ log y = cos x log (log x)

⇒ \(\frac{d }{d x}\)logy = \(\frac{d }{d x}\)cosx log(logx)

Question 17.

Find the interval in which the function f given f(x) = 2x^{2} – 3x is strictly increasing.

Answer:

f(x) = 2x^{2} – 3x

diff wrt x

f^{1} (x) = 4x – 3

f^{1} (x) = 0 ⇒ 4x – 3 = 0

or 4x = 3

⇒ x = 3/4

∴ Point x = 3/4 divides the real line into two disjoint intervals (- ∞, 3/4) and (3/4, ∞) In the interval (3/4, ∞), f^{1}(x) = 4x – 3 > 0

∴ f is strictly increasing in (3/4, ∞)

Question 18.

Find \(\int \frac{\left(x^{4}-x\right)^{\frac{1}{4}}}{x^{5}} d x\)

Answer:

Question 19.

Integrate x sec^{2} x with respect to x.

Answer:

\(\int x \sec ^{2} x d x=x \int \sec ^{2} x d x-\int\left(\int \sec ^{2} x\right) \cdot \frac{d}{d x}(x) \cdot d x\)

= x. tan x – \(\int \tan x \cdot 1 \cdot d x\)

= x . tanx – log |sec x| + c

Question 20.

Find the order and degree (if defined) of the differential equation :

y^{111} – y^{2} + e^{3} = 0

Answer:

Order = 3, Degree = not defined.

Question 21.

If \(\overrightarrow{\mathrm{a}}\) is a unit vector and \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 8, then find \(|\overrightarrow{\mathrm{x}}|\) .

Answer:

Question 22.

Find the area of the parallelogram whose adjacent sides are given by the vectors \(\overrightarrow{\mathrm{a}}\) = 3î + ĵ + 4 k̂ and \(\overrightarrow{\mathrm{b}}\) = î – ĵ + k̂

Answer:

Area of parallelogram = \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|\)

Question 23.

Find the angle between the pair of lines \(\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}\) and \(\frac{x+1}{1}=\frac{y-1}{1}=\frac{z-5}{2}\)

Answer:

Given

\(\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}\)

Here \(\overrightarrow{\mathrm{b}}_{1}\) = 3î + 5ĵ + 4k̂

and \(\frac{x+1}{1}=\frac{y-1}{1}=\frac{z-5}{2}\) ⇒ î + ĵ + 2k̂

Question 24.

The probability distribution of random variable X is as follows :

Answer:

Here E(X) = Σx_{i} P(x_{i})

\(=0 \times \frac{188}{221}+1 \times \frac{32}{221}+2 \times \frac{1}{221}=\frac{32+2}{221}=\frac{34}{221}\)

Part-C

Answer any TEN questions: (10 × 3 = 30)

Question 25.

Determine whether the relation R in the set A = {1, 2, 3, …., 13, 14} defined as R = {(x,y); 3x – y = 0} is reflexive, symmetric and transitive.

Answer:

Given A = {1,2,3, ….,13, 14}

R = {(x, y) : 3x – y =0}

∴ R= {1,3), (2, 6), (3, 9), (4, 12)}

Here (1, 1) ∉ R, (2, 2) ∉ R

R is not reflexive

Here (1, 3) ∈ R but (3, 1) ∉ R

∴ R is not symmetric

Here (1,3) and (3, 9)6 ∈ R but (1,9) ∉ R

∴ R is not transitive

∴ R is not an equivalence relation.

Question 26.

Solve for x, if tan^{-1} 2x + tan^{-1}3x = \(\frac{\pi}{4}\), x > 0

Answer:

Given tan^{-1}(2x) + tan^{-1}(3x) = \(\frac{\pi}{4}\)

tan^{-1} \(\left(\frac{2 x+3 x}{1-2 x \cdot 3 x}\right)\) = \(\frac{\pi}{4}\)

\(\frac{5 x}{1-6 x^{2}}\) = tan \(\frac{\pi}{4}\)

⇒ \(\frac{5 x}{1-6 x^{2}}\) = 1 ⇒ 5x = 1 – 6x^{2}

∴ 6x^{2} + 5x – 1 = 0

6x^{2} + 6x – x – 1 =0

6x(x+1) -1(x+1) = 0 or (x+1) (6x – 1)= 0

⇒ x = -1 or x = \(\frac { 1 }{ 6 }\)

Question 27.

By using elementary transformations, find the inverse of the matrix

Answer:

Question 28.

If x = a(θ + sinθ), y = a( 1 – cosθ) then show that \(\frac{d y}{d x}\) = tan \(\left(\frac{\theta}{2}\right)\).

Answer:

we have

\(\frac{d x}{d \theta}\) = a(1 + cosθ), \(\frac{d y}{d \theta}\) = a(sin θ)

Question 29.

Verify mean value theorem for the function f(x) = x^{2} – 4x – 3 in the interval [1, 4].

Answer:

f(x) = x^{2} – 4x – 3 interval [a, b] = [1,4]

Here f is a polynomial function hence continuous in [1, 4] and differentiable in (1,4)

f(x) = x^{2} – 4x – 3 (diff wrt x)

∴ f^{1}(x) = 2x – 4 1

Also f(1) = 1^{2} – 4 – 3 = -6

f(4) = 4^{2} – 4(4) – 3 = -3

∴ From MVT ∃ c ∈ (a,b) such that

f^{1} (c) = \(\frac{f(b)-f(a)}{b-a}=\frac{-3-(-6)}{4-1}\)

f^{1} (c) = \(\frac { 3 }{ 3 }\) = 1

f^{1} (c) = 2c – 4

∴ 2c – 4 = 1

or 2c = 5

c = \(\frac { 5 }{ 2 }\) ∈ (1, 4)

Question 30.

Find the point, at which the tangent to the curve y = \(\sqrt{4 x-3}-1\) has its slope \(\frac { 2 }{ 3 }\)

Answer:

y = \(\sqrt{4 x-3}-1\) (diff wrt x)

\(\frac{d y}{d x}=\frac{1}{2 \sqrt{4 x-3}} \cdot 4\)

Given slope of tangent = \(\frac{d y}{d x}=\frac{2}{3}\)

∴ \(\frac{2}{3} =\frac{1}{2 \sqrt{4 x-3}} .4\)

\(\frac{4}{3} =\frac{4}{\sqrt{4 x-3}} \Rightarrow \sqrt{4 x-3}=3\)

\(\sqrt{4 x-3} = 3\) (squaring)

4x – 3 = 9

4x = 12

0r x = 3

∴ y = \(\sqrt{4(3)-3-1}\)

= \(\sqrt{9}\) – 1 = 3 – 1 = 2

∴ point = (3, 2)

Question 31.

Find \(\int \frac{x}{(x+1)(x+2)} d x\).

Answer:

Question 32.

Evaluate \(\int_{0}^{2}\left(x^{2}+1\right) d x\). as a limit of a sum.

Answer:

\(\int_{0}^{2}\left(x^{2}+1\right) d x\) Here a = 0 ; b = 2

f (x) = x^{2} + 1

Question 33.

Find the area of the region bounded by y^{2} = 9x, x = 2, x = 4 and the x – axis in the first quadrant.

Answer:

Question 34.

If y = ae^{3x} + be^{-2x} represents family of curves, where a and b are arbitrary constants. Form the differential equation.

Answer:

Given, y = ae^{3x} + be^{-2x} …. (1)

y’ = (ae^{3x})3 – (be^{-2x}) 2

y’ = 3a e^{3x} – 2be^{-2x} …. (2)

From (1) and (2)

y = ae^{3x }+ be^{-2x} × 3

y’ = 3a e^{3x} – 2b e^{-2x} × 1

Diff

⇒ 3y’ – y” = 5be^{-2x}

3y’ – y”= -10be^{-2x} …. (4)

From (3) & (4)

3y’ – y” = (-2) (5be^{-2x})

3y’ – y” = -2[3y – y’]

3y’- y” = -6y + 2y’

y” – y’ – 6y= 0.

Question 35.

Show that the position vector of the point P, which divides the line joining the points A and B having position vectors \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\) internally in the ratio

\(\mathbf{m}: \mathbf{n} \text { is } \frac{\mathbf{m} \overrightarrow{\mathbf{b}}+\mathbf{n} \overrightarrow{\mathbf{a}}}{\mathbf{m}+\mathbf{n}}\)

Answer:

Here P divides AB internally in the ratio m : n

∴ p divides \(\overrightarrow{\mathrm{AB}}\) such that

\(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{m}}{\mathrm{n}}\) or n\(\overrightarrow{\mathrm{AP}}\) = m\(\overrightarrow{\mathrm{PB}}\)

Where m and n are positive scalars

Here \(\overrightarrow{\mathrm{r}}, \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}\) are position vectors of \(\overrightarrow{\mathrm{OP}}\)

\(\overrightarrow{\mathrm{OA}}\) an \(\overrightarrow{\mathrm{OB}}\) respectively

∴ From the figure we have,

Question 36.

Prove that \([\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}, \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}}]\) = 2\([\vec{a}, \vec{b}, \vec{c}]\)

Answer:

LHS

Question 37.

Find the shortest distance between the lines \(\overrightarrow{\mathbf{r}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\lambda(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\) \(\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}+\mu(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\)

Answer:

2nd PUC Maths Previous Year Question Paper June 2019 27

Question 38.

Bag I contains 3 red and 4 black balls. White Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find

Answer:

Let A be the event of choosing the bag I

B be the event of choosing the bag II

and C be the event of drawing a red ball

∴ P(A) = P(B) = \(\frac { 1 }{ 2 }\)

Also p(C/A) = \(\frac { 3 }{ 7 }\) and P(C/B) = \(\frac { 5 }{ 11 }\)

∴ P(C)=P(A).P(C/A) +P(B).P(C/B)

\(=\frac{1}{2} \times \frac{3}{7}+\frac{1}{2} \times \frac{5}{11}=\frac{3}{14}+\frac{5}{22}=\frac{33+35}{154}\)

P(C) = \(\frac { 68 }{ 154 }\)

By Baye’s Theorem,

P(B/C) = \(\frac { 35 }{ 68 }\)

Part-D

Answer any SIX questions: (6 × 5 = 30)

Question 39.

Prove that the function f = R → R defined by f(x) = 4x + 3, is invertible and find the inverse of f.

Answer:

Let f(x_{1}) = f(x_{2})= 4x_{1} + 3 ⇒ 4x_{2} + 3

⇒ 4x_{1} = 4x_{2} ⇒ x_{1} = x_{1}

f is onto.

Let y ∈ R∃ x ∈ R such that f(x) = y

4x + 3 = y \(x=\frac{y-3}{4} \in R\)

∴ f is onto ∴ f is bijection

∴ f is invertible

Define a function g : R → R by

g(y) = \(\frac{y-3}{4}\)

(fog) (y) = f(g(y)) = f\(\frac{y-3}{4}\)

= 4\(\frac{y-3}{4}\)+3 = y – 3 + 3 = y ∴ (fog) (y) = y fog = I

(gof) (x) = g(f(x)) = g (4x + 3)

\(=\frac{4 x+3-3}{4}=\frac{4 x}{4}=x\)

∴ (gof) (x) = x ∴gof = I

∴ f^{-1}(y) = g(y) = \(\frac{y-3}{4}\) ∀ y ∈ R

Question 40.

If

Calculate AC, BC and (A + B)C.

Answer:

Question 41.

Solve the following system of linear equations by matrix method:

3x – 2y + 3z = 8

2x + y – z = 1

4x – 3y + 2z = 4

Answer:

The system of equations can be written in the form A X = B where

|A| = 3 (2 – 3) + 2(4 + 4) + 3 (-6 -4) = – 17

Hence, A is non singular and so its inverse exists. Now

A_{11} = -1, A_{12} = -8 A_{13} = -10

A_{21} = -5, A_{22} = -6 A_{23} = +1

A_{31} = -1, A_{32} = +9 A_{33} = +7

A^{-1} = \(\frac{1}{|\mathrm{A}|}\) adj A

substituting in (1)

Question 42.

If y = 3 cos (log_{e}x) + 4 sin (log_{e}x) show that x^{2}y_{2} + xy_{1} + y = 0.

Answer:

y = 3 cos (log_{e}x) + 4sin (log_{e}x)

Diff wrt x

y_{1} = 3(-sin(logx).\(\frac { 1 }{ x }\)) + 4 cos (log x). \(\frac { 1 }{ 2 }\)

xy_{1} = -3 sin (logx) + 4 cos(logx)

diff wrt x

xy_{2} y_{1}. 1 = -3cos(logx).\(\frac { 1 }{ x }\) + 4(-sin(logx)).\(\frac { 1 }{ x }\)

xy_{2} + y_{1} = \(\frac { 1 }{ 2 }\)-(3cos(logx) + 4sin(logx))

x^{2}y_{2} + xy_{1} = -(3cos(logx) + 4sin(logx))

x^{2}y_{2} + xy_{1} = -y

∴ x^{2}y_{2}+ xyj_{1}+ y = 0

Question 43.

Sand is pouring from a pipe at the rate of 12cm3/sec. The falling sand forms a cone on the ground in such a way that the height of the cone is always one – sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Answer:

Volume, V = \(\frac{\pi r^{2} h}{3}\)

given; h = \(\frac { 1 }{ 6 }\) r (or) r = 6h

∴ v = \(\frac{\pi}{3}\) × (6h)^{2} ×h = 12πh^{3}

v = 12πh^{3}

diff w.r.t ‘t’

Question 44.

Find the integral of \(\sqrt{x^{2}+a^{2}}\) with respect to x, and hence find \(\int \sqrt{x^{2}+2 x+5} d\)

Answer:

Question 45.

Using integration find the area of the region in the first quadrant enclosed by the x – axis, the line y = x, ad circle x^{2} + y^{2} = 32

Answer:

x^{2} + y^{2} = 32 and centre (0,0)

∴ radius = r = \(\sqrt{32}\) = \(4 \sqrt{2}\)

Equation are y = x ….(1) and x^{2} + y^{2} = 32 …..(2)

∴x^{2} + x^{2} = 32 (usinig (1) and (2))

∴ 2x^{2} = 32 ⇒ x^{2} = 16 or x = ±

since region is in first quadrant

x = 4 ⇒ y = 4

∴ Required Area = Area OBM + Area MBA

Question 46.

Find the general solution of the differential equation (x+ y)\(\frac{d y}{d x}\) = 1

Answer:

(x + y)\(\frac{d y}{d x}\) = l dx

(x + y)dy = dx or \(\frac{d y}{d x}\) = x + y

∴ \(\frac{d y}{d x}\) -x = y which i s in the form

\(\frac{d y}{d x}\) + Px =Q dy

Here P = -1; Q = y

∴ Integrating Factor = IF = \(e^{\int P d y}\)

\(=e^{\int(-1) d y}=e^{-y}\)

∴ General solution is

x(IF) = \(\int y \cdot(I F) d y+C\) (Using integration by parts)

= y(-e^{-y}) – \(\int-\left(e^{-y}\right)\).1.dy + C

= -ye^{-y} +\(\int e^{-y} d y\) + C ÷ e^{-y}

xe^{-y} = -ye^{-y} – e^{-y} + C

⇒ x = -y-l+C.e^{y}

Or x + y + 1 = C.e^{y}

Question 47.

Derive the equation of a plane in normal form both in the vector and Cartesian form.

Answer:

Consider a plane whose perpendicular distance from the origin is d. where d ≠ 0.

This is the Required Equation of a plane m normal form.

To find P and Q :

x^{2} = 4y and x =4y – 2

=>(4y – 2)^{2} = 4y ⇒ 16y^{2} + 4 – 16 y=4 y

⇒ 16y^{2} – 20y +4 = 0

⇒ (y – 1) (16y – 4) = 0

⇒ y = 1 (or) y = \(\frac { 1 }{ 4 }\)

if y = 1 ; x = 2 ⇒ 2 (2,1)

and y = \(\frac { 1 }{ 4 }\) ; x = -1 ⇒ (-1, \(\frac { 1 }{ 4 }\))

Thus the points of intersection are

(2,1) I(-1 .\(\frac { 1 }{ 4 }\))

∴ Area of the Region O B Q C = Area of BCQ + Area of QCO

Question 48.

A die is thrown 6 times. If ‘getting an odd number’ is success, what is the probability of

a) 5 successes

b) at least 5 successes

c) at most 5 Successes

Answer:

Let P = probability of getting an odd number = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

q = 1 – p = 1\(\frac { -1 }{ 2 }\) = \(\frac { 1 }{ 2 }\)

& n =6

we have P(x) = ^{n}c_{x} P^{x} q^{n-x}

Part-E

Answer any ONE question: (1 × 10 = 10)

Question 49.

a) Prove that \(\int_{0}^{b} f(x) \cdot d x=\int_{a}^{b} f(a+b-x) d x\) and hence. Evaluate \(\int_{\pi / 3}^{\pi / 3} \frac{1}{1+\sqrt{\tan x}} d x\) (June – 2014,Q.No.49a)

Solution :

let t = a + b – x then dt = -dx

when x =a and x = b

t = b t = a

f(x) is continuous, find the value of k.

Question 50.

(a) A factory manufactures two types of screws, A and B. Each type of Screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at most 4 hours on any day. The manufacture can sell a package of screws A at a profit of Rs.7 and seres B at a profit of Rs.10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.

Answer:

∴ Maximise Z = 7x + 10y subject to constraints

4x + 6y ≤ 240

or 2x + 3y ≤ 120 …… (1)

and 6x + 3y ≤ 240

or 2x + y ≤ 80 …… (2)

and x, y ≥ 0

Converting inequalities to equalities

2x + 3y= 120

2x + y = 80

Plot the graph

Maximum profit is Rs.410 which occurs at point B (30, 20)

∴Manufacturer should manufacture 30 packages of screws A and 20 packages of screws B.

Answer:

= (1 + x + x^{2})(1 – x^{2}) [1(1 + x + x^{2}) – 0 + 0] (Expanding along c_{1})

= (1 + x + x^{2}) (1 + x + x^{2}) (1 – x^{2})

= [(1 + x + x^{2}) (1 – x^{2})]^{2} = (1^{3} – x^{3})^{2} (using a^{3} – b^{3}

= (a – b) (a^{2} + b^{2} +ab))

= (1 – x^{3})^{2} (RHS)